Question

Create a rational function that has the following characteristics: crosses the \(x\) -axis at \(3 ;\) touches the \(x\) -axis at -2 ; one vertical asymptote, \(x=1\); and one horizontal asymptote, \(y=2\). Give your rational function to a fellow classmate and ask for a written critique of your rational function. Create a rational function that has the following characteristics: crosses the \(x\) -axis at \(2 ;\) touches the \(x\) -axis at \(-1 ;\) one vertical asymptote at \(x=-5\) and another at \(x=6 ;\) and one horizontal asymptote, \(y=3 .\) Compare your function to a fellow classmate's. How do they differ? What are their similarities?

Short answer

First function: \ \frac{2(x-3)(x+2)^2}{(x-1)} \ . Second function: \ \frac{3(x-2)(x+1)^2}{(x+5)(x-6)} \.

Step-by-step solution

Understand the problem requirements

We need to create a rational function with specific characteristics: crossing the x-axis, touching the x-axis, vertical asymptotes, and a horizontal asymptote. Note the requirements carefully for each function.

Determine roots and behavior at the x-axis

The function crosses the x-axis at 3 and 2. Therefore, it will have factors \(x - 3\) and \(x - 2\). It touches the x-axis at -2 and -1, meaning squared factors \(x + 2\)^2 and \(x + 1\)^2 will be used.

Identify vertical asymptotes

For the first function, the vertical asymptote is at \(x = 1\) and for the second function, the vertical asymptotes are at \(x = -5\) and \(x = 6\). These give us factors in the denominator such as \(x - 1\), \(x + 5\), and \(x - 6\).

Determine the horizontal asymptote

Horizontal asymptotes are related to the behavior of the function as \ x \ approaches infinity. For the first function, \(y = 2\), implying the degree of numerator and denominator is the same with coefficients leading to 2. Similarly, for the second function, \(y = 3\).

Construct the rational function - First Function

Combine these details: the numerator will be \(k(x - 3)(x + 2)^2\) divided by \( x - 1\). To make the horizontal asymptote \(y = 2\), k must adjust to ensure the leading coefficient ratio is 2, thus \[\frac{2(x-3)(x+2)^2}{(x-1)}. \]

Construct the rational function - Second Function

Combine details: \ k(x - 2)(x + 1)^2 \ divided by \ (x + 5)(x - 6) \. Adjust k to ensure the leading coefficient ratio is 3 for the horizontal asymptote, thus \[\frac{3(x-2)(x+1)^2}{(x+5)(x-6)}. \]

Comparison

Compare each function structurally: one has a single vertical asymptote and simpler factors, while the other has two vertical asymptotes with more complex factors in the denominator.

Key concepts

Understanding Vertical Asymptotes

A vertical asymptote represents a value of x where the rational function shoots up to positive or negative infinity. This happens because the denominator of the function approaches zero, making the function undefined at that point. For example, consider the vertical asymptote at x=1 for the first rational function: \[\frac{2(x-3)(x+2)^2}{x-1}\]. As x gets closer to 1, the denominator approaches zero, causing the whole function to increase or decrease without bound. Remember that the numerator does not influence the vertical asymptote directly; it determines whether the function goes to positive or negative infinity near the asymptote.

To identify vertical asymptotes:

• Examine the denominator of the rational function.
• Find the values of x where the denominator is zero.
• Each of these values corresponds to a vertical asymptote.

Exploring Horizontal Asymptotes

Horizontal asymptotes describe the behavior of the rational function as x approaches infinity or negative infinity. They show the value that the y-coordinate of the function gets closer to. In our example rational functions, the horizontal asymptote is determined by the leading coefficients of the numerator and denominator when their degrees are the same. For the first function, the horizontal asymptote is at y=2 \[2 \frac{(x-3)(x+2)^2}{x-1}\]. The leading coefficient of the numerator and denominator is used to determine this. If the degrees of the denominator and numerator are different, the asymptote may be at zero or non-existent.

Key points about horizontal asymptotes:

• When the degrees are equal, the horizontal asymptote is the ratio of leading coefficients.
• If the numerator's degree is less, the horizontal asymptote is at y=0.
• If the numerator's degree is greater, there is no horizontal asymptote.

Understanding Roots of a Function

The roots of a function are the values of x where the function crosses or touches the x-axis (the y-value is zero). For the exercise, the provided function has roots at x=3 and x=-2, meaning the function equals zero when these values are substituted for x. This is indicated by factors in the numerator like (x-3) and (x+2)^2. The square means the function touches the x-axis at x=-2 without crossing it, as it bounces back.

To find and understand the roots:

• Identify the factors of the numerator.
• Set each factor to zero and solve for x.
• These roots will be the x-intercepts of the function.

Behavior at Infinity

As x approaches infinity or negative infinity, the behavior of the rational function is primarily determined by the highest degree terms in the numerator and denominator. For the provided functions, \[\frac{2(x-3)(x+2)^2}{x-1}\] and \[\frac{3(x-2)(x+1)^2}{(x+5)(x-6)}\], examine the leading coefficient in both the numerator and denominator.

This behavior dictates the horizontal asymptote's location. In our example, the first function's leading terms' ratio is 2, leading to a horizontal asymptote at y=2. Similarly, for the second function, it's y=3.

For a deeper look:

• Identify the highest degree terms of both the numerator and denominator.
• Compute the ratio of their coefficients.
• This ratio will determine the horizontal asymptote's value.